python获取字符串中的email

 调用re库，通过使用compile、findall获取字符串中的email

import re
email=re.compile(r'[a-z0-9\-\.]+@[0-9a-z\-\.]+')
emailtest=r'adfasldfjdsl fan02@163.com'
emailset=set()
for em in email.findall(emailtest):
    emailset.add(em)
for em1 in sorted(emailset):
    print(em1)

修改：

通过调用函数，并动态地为字符串赋值

import re
def emailre(teststr):
    email=re.compile(r'[a-z0-9\-\.]+@[0-9a-z\-\.]+')
    emailset=set()  #列表
    for em in email.findall(teststr):
        emailset.add(em)
    for eml in sorted(emailset):
        print(eml)
emailtest='sdfdsgf asd03@162.com'
emailre(emailtest)
#或
strtest=r'sdgfsg abc@163.com'
emailre(strtest)

运行结果：

补充：

从指定的字符串中提取Email：

  /**
   * 从指定的字符串中提取Email
   * content 指定的字符串
   */
  public static String parse(String content) {
  String email = null;
  if (content==null || content.length()<1) {
 return email;
  }
 //找出含有@
 int beginPos;
 int i;
String token = "@";
String preHalf="";
 String sufHalf = "";

beginPos = content.indexOf(token);
 if (beginPos>-1) {
 //前项扫描
 String s = null;
  i= beginPos;
 while(i>0) {
s = content.substring(i-1,i);
 if (isLetter(s))
    preHalf = s+preHalf;
  else
    break;
  i--;
  }
 //后项扫描
  i= beginPos+1;
  while( i<content.length()) {
   s = content.substring(i,i+1);
   if (isLetter(s))
    sufHalf =  sufHalf +s;
   else
    break;
    i++;  
   }
  //判断合法性
  email = preHalf + "@" + sufHalf;
   if (isEmail(email)) {
   return email;
 }
  }
return null;
}

 /**
 * 判断是不是合法Email
* email Email地址
*/
public static boolean isEmail(String email) {
 try {
   if (email==null || email.length()<1 || email.length()>256) {
   return false;
 }
   
String check = "^([0-9a-zA-Z]+[_.0-9a-zA-Z-]+)@([a-zA-Z0-9-]+[.])+([a-zA-Z]{2,3})$";
 Pattern regex = Pattern.compile(check);
 Matcher matcher = regex.matcher(email);
  boolean isMatched = matcher.matches();
 if(isMatched) {
   return true;
 } else {
  return false;
 }
 } catch (RuntimeException e) {
 return false;
  } 
 }
 
 /**
  * 判断是不是合法字符
 * c 要判断的字符
 */
public static boolean isLetter(String c) {
 boolean result = false;
 if (c==null || c.length()<0) {
  return false;
 }
 //a-z 
 if (c.compareToIgnoreCase("a")>=0 && c.compareToIgnoreCase("z")<=0) {
  return true;
 }
 //0-9
 if (c.compareToIgnoreCase("0")>=0 && c.compareToIgnoreCase("9")<=0) {
 return true;
 }
 //. - _
 if (c.equals(".") || c.equals("-") || c.equals("_") ) {
 return true;
}
 return result; 
 }

来源：https://blog.csdn.net/aigo_2021/article/details/123859764